Question-223560

Question Number 223560 by fantastic last updated on 29/Jul/25

Commented by fantastic last updated on 29/Jul/25

$${area}=? \\ $$

Commented by fantastic last updated on 29/Jul/25

keep in mind that it is a grade 9 question and trigonomerty and any higher level method is not allowed

$${keep}\:{in}\:{mind}\:{that}\:{it}\:{is}\:{a}\:{grade}\:\mathrm{9} \\ $$$${question}\:{and}\:{trigonomerty}\:{and} \\ $$$${any}\:{higher}\:{level}\:{method}\:{is}\:{not}\:{allowed} \\ $$

Commented by AgniMath last updated on 29/Jul/25

Are you west bengal board candidate class 9?

$${Are}\:{you}\:{west}\:{bengal}\:{board}\:{candidate} \\ $$$${class}\:\mathrm{9}? \\ $$

Commented by AgniMath last updated on 29/Jul/25

Commented by mr W last updated on 29/Jul/25

Commented by fantastic last updated on 29/Jul/25

$${yes}. \\ $$$${How}\:{did}\:{you}\:{know}?? \\ $$

Commented by fantastic last updated on 29/Jul/25

$${thanks}\:{sir} \\ $$

Commented by fantastic last updated on 29/Jul/25

$${Thanks}\:{sir} \\ $$$$ \\ $$

Commented by AgniMath last updated on 30/Jul/25

I read in class 9 west bengal board before (now in 11th). I remembered this question.

$${I}\:{read}\:{in}\:{class}\:\mathrm{9}\:{west}\:{bengal}\:{board}\: \\ $$$${before}\:\left({now}\:{in}\:\mathrm{11}{th}\right).\:{I}\:{remembered}\:{this} \\ $$$${question}. \\ $$

Commented by fantastic last updated on 30/Jul/25

$${thanks}\:{sir}. \\ $$

Commented by ajfour last updated on 30/Jul/25

https://youtu.be/x1KJRC4GqpI?si=pilhm24h2cFqVOig

Commented by mr W last updated on 30/Jul/25

$${an}\:{other}\:{way}\:{as}\:{in}\:{video}: \\ $$

Commented by mr W last updated on 30/Jul/25

Commented by mr W last updated on 31/Jul/25

(a+b+(√(a^2 +b^2 )))r=ab ⇒r=((ab)/(a+b+(√(a^2 +b^2 ))))=(1/( (√((1/a^2 )+(1/b^2 )))+(1/a)+(1/b))) (r_a /r)=((b+2r_a )/b) ⇒r_a =(1/((1/r)−(2/b)))=(1/( (√((1/a^2 )+(1/b^2 )))+(1/a)−(1/b))) similarly ⇒r_b =(1/((1/r)−(2/a)))=(1/( (√((1/a^2 )+(1/b^2 )))−(1/a)+(1/b))) example in video: a=5, b=(4/3)×5=((20)/3) r=((5×((20)/3))/(5+((20)/3)+(√(5^2 +(((20)/3))^2 ))))=(5/3) (=r) r_a =(1/((3/5)−((2×3)/(20))))=((10)/3) (=R) r_b =(1/((3/5)−(2/5)))=5 (=ρ)

$$\left({a}+{b}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right){r}={ab} \\ $$$$\Rightarrow{r}=\frac{{ab}}{{a}+{b}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }}+\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}} \\ $$$$\frac{{r}_{{a}} }{{r}}=\frac{{b}+\mathrm{2}{r}_{{a}} }{{b}} \\ $$$$\Rightarrow{r}_{{a}} =\frac{\mathrm{1}}{\frac{\mathrm{1}}{{r}}−\frac{\mathrm{2}}{{b}}}=\frac{\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }}+\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{{b}}} \\ $$$${similarly} \\ $$$$\Rightarrow{r}_{{b}} =\frac{\mathrm{1}}{\frac{\mathrm{1}}{{r}}−\frac{\mathrm{2}}{{a}}}=\frac{\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }}−\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}} \\ $$$${example}\:{in}\:{video}:\:{a}=\mathrm{5},\:{b}=\frac{\mathrm{4}}{\mathrm{3}}×\mathrm{5}=\frac{\mathrm{20}}{\mathrm{3}} \\ $$$${r}=\frac{\mathrm{5}×\frac{\mathrm{20}}{\mathrm{3}}}{\mathrm{5}+\frac{\mathrm{20}}{\mathrm{3}}+\sqrt{\mathrm{5}^{\mathrm{2}} +\left(\frac{\mathrm{20}}{\mathrm{3}}\right)^{\mathrm{2}} }}=\frac{\mathrm{5}}{\mathrm{3}}\:\:\:\:\left(={r}\right) \\ $$$${r}_{{a}} =\frac{\mathrm{1}}{\frac{\mathrm{3}}{\mathrm{5}}−\frac{\mathrm{2}×\mathrm{3}}{\mathrm{20}}}=\frac{\mathrm{10}}{\mathrm{3}}\:\:\:\:\:\:\:\:\left(={R}\right) \\ $$$${r}_{{b}} =\frac{\mathrm{1}}{\frac{\mathrm{3}}{\mathrm{5}}−\frac{\mathrm{2}}{\mathrm{5}}}=\mathrm{5}\:\:\:\:\:\:\:\:\:\:\left(=\rho\right) \\ $$

Commented by ajfour last updated on 31/Jul/25

$${Thank}\:{you}\:{sir}!\:{keen}\:{notice}. \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply